1. **Stating the problem:** We have a system of linear equations: $$ \begin{cases} x_1 + x_2 + 3x_3 = -1 \\ 3x_1 + 2x_2 - 2x_3 = 1 \end{cases} $$ and a vector solution expressed as: $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -3 \\ -2 \\ 0 \end{bmatrix} + s \begin{bmatrix} -14 \\ -11 \\ 1 \end{bmatrix} $$ 2. **Goal:** Verify that this vector expression is a solution to the system and understand the solution structure. 3. **Step 1: Understand the solution form** The solution is given as a particular solution plus a scalar multiple of a vector: $$ \mathbf{x} = \mathbf{p} + s \mathbf{v} $$ where $$ \mathbf{p} = \begin{bmatrix} -3 \\ -2 \\ 0 \end{bmatrix}, \quad \mathbf{v} = \begin{bmatrix} -14 \\ -11 \\ 1 \end{bmatrix} $$ 4. **Step 2: Verify the particular solution satisfies the system** Substitute $x_1 = -3$, $x_2 = -2$, $x_3 = 0$ into the equations: - First equation: $$ -3 + (-2) + 3 \times 0 = -3 - 2 + 0 = -5 \neq -1 $$ This suggests the vector given is not a solution to the system as stated, so let's check the system carefully. 5. **Step 3: Re-examine the system and solution** Since the particular solution does not satisfy the system, likely the system is incomplete or the solution corresponds to a different system. However, the problem only provides two equations for three variables, so the system is underdetermined and has infinitely many solutions. 6. **Step 4: Express the system in matrix form and solve** Matrix form: $$ \begin{bmatrix} 1 & 1 & 3 \\ 3 & 2 & -2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} $$ 7. **Step 5: Use the given solution form to check if it satisfies the system** Substitute $\mathbf{x} = \mathbf{p} + s \mathbf{v}$: $$ A(\mathbf{p} + s \mathbf{v}) = A\mathbf{p} + s A\mathbf{v} = \mathbf{b} $$ Check $A\mathbf{p}$: $$ \begin{bmatrix} 1 & 1 & 3 \\ 3 & 2 & -2 \end{bmatrix} \begin{bmatrix} -3 \\ -2 \\ 0 \end{bmatrix} = \begin{bmatrix} -3 - 2 + 0 \\ -9 - 4 + 0 \end{bmatrix} = \begin{bmatrix} -5 \\ -13 \end{bmatrix} $$ Check $A\mathbf{v}$: $$ \begin{bmatrix} 1 & 1 & 3 \\ 3 & 2 & -2 \end{bmatrix} \begin{bmatrix} -14 \\ -11 \\ 1 \end{bmatrix} = \begin{bmatrix} -14 - 11 + 3 \\ -42 - 22 - 2 \end{bmatrix} = \begin{bmatrix} -22 \\ -66 \end{bmatrix} $$ 8. **Step 6: Since $A\mathbf{p} \neq \mathbf{b}$, the given vector $\mathbf{p}$ is not a particular solution.** 9. **Step 7: Find the correct particular solution** Let $x_3 = t$ (free variable). Then from the first equation: $$ x_1 + x_2 = -1 - 3t $$ From the second equation: $$ 3x_1 + 2x_2 = 1 + 2t $$ 10. **Step 8: Solve for $x_1$ and $x_2$ in terms of $t$** Multiply the first equation by 2: $$ 2x_1 + 2x_2 = -2 - 6t $$ Subtract from the second equation: $$ (3x_1 + 2x_2) - (2x_1 + 2x_2) = (1 + 2t) - (-2 - 6t) $$ Simplify: $$ x_1 = 3 + 8t $$ Substitute back to find $x_2$: $$ 3 + 8t + x_2 = -1 - 3t \implies x_2 = -4 - 11t $$ 11. **Step 9: Write the general solution vector** $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ -4 \\ 0 \end{bmatrix} + t \begin{bmatrix} 8 \\ -11 \\ 1 \end{bmatrix} $$ 12. **Step 10: Final answer** The solution set is: $$ \boxed{\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ -4 \\ 0 \end{bmatrix} + t \begin{bmatrix} 8 \\ -11 \\ 1 \end{bmatrix}, \quad t \in \mathbb{R}} $$ This means the system has infinitely many solutions parameterized by $t$.